24-Pet-A6 Well Logging and Formation Evaluation · May 2017
Question 7 of 8
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2017 – 98-Pet-A6, Reservoir Mechanics (3 hours, open book, non-communicating calculator permitted). The exam consists of EIGHT questions; candidates respond to 100 marks of their choice out of 120. All eight questions are answered in full below.
Reference texts: Craft, B.C. and Hawkins, M. (rev. Terry & Rogers), Applied Petroleum Reservoir Engineering, 3rd ed.; Ahmed, T., Reservoir Engineering Handbook, 5th ed.; Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.
If doubt exists..." "2. This is an OPEN BOOK EXAM..." "3. The exam consists of EIGHT (8) questions..." "4. Clarity and organization...") immediately above Question 1. every real "Question N" heading on this paper parses cleanly (checked by direct read of the source), so it does not affect the answers below.
Given. A linear "shoestring" waterflood with explicit relative-permeability functions (no chart reading required).
Given data
$W\times h\times L$
300 × 20 × 1,000 ft
$\phi$
0.15
$S_{wi}$
0.30
$S_{orw}$
0.20
$\mu_o,\ \mu_w$
2.0, 1.0 cp
$q_{inj}$
350 bbl/day
Fig. Q7 – fractional-flow curve $f_w(S_w)$ with the Welge tangent from $S_{wi}$, marking the shock front $S_{wf}$ and the producing-well saturation $S_{w2}=0.66$.
Find. (a) $N_p$ at $S_{w2}=0.66$; (b) $W_p$ at the same condition; (c) areal sweep after 25,000 bbl injected.
Approach. Build the fractional-flow curve from the given relative-permeability functions, apply Welge's method (average saturation behind the front from the tangent slope at $S_{w2}$) for parts (a)–(b), and use the Buckley–Leverett frontal-advance relation for the shock front to answer the pre-breakthrough sweep in part (c).
Fractional-flow function (no gravity, horizontal bed).
$$f_w(S_w)=\frac{1}{1+\dfrac{\mu_w}{\mu_o}\dfrac{k_{ro}}{k_{rw}}}=\frac{1}{1+\dfrac12\dfrac{(1-S_{wD})^3}{S_{wD}^4}}$$
At $S_{w2}=0.66$: $S_{wD}=\dfrac{0.66-0.30}{1-0.20-0.30}=0.72$, giving $f_{w2}=0.9608$ and (by numerical differentiation of $f_w$) $\left.\dfrac{df_w}{dS_w}\right|_{0.66}=1.227$.
Average saturation behind the front (Welge).
$$\bar S_w=S_{w2}+\frac{1-f_{w2}}{\left(df_w/dS_w\right)_{S_{w2}}}=0.66+\frac{1-0.9608}{1.227}$$
$$\boxed{\bar S_w\approx0.692}$$
Cumulative oil production at $S_{w2}=0.66$.
$$N_p=V_p(\bar S_w-S_{wi})=160{,}285(0.692-0.30)$$
$$\boxed{N_p\approx 62{,}830\text{ STB}}$$
Cumulative water production at the same condition. Material balance (water injected = water retained + water produced), using $Q_{iD}=1/f_w'(S_{w2})$ for the dimensionless cumulative injection:
$$W_p=V_p\!\left[\frac{f_{w2}}{\left(df_w/dS_w\right)_{S_{w2}}}-(S_{w2}-S_{wi})\right]=160{,}285\left[\frac{0.9608}{1.227}-0.36\right]$$
$$\boxed{W_p\approx 67{,}828\text{ STB}}$$
Shock-front saturation and areal sweep after 25,000 bbl injected. The shock (leading front) saturation $S_{wf}$ is found from the Welge tangent condition $f_w(S_{wf})/(S_{wf}-S_{wi})=(df_w/dS_w)_{S_{wf}}$, solved numerically: $S_{wf}\approx0.626$, giving $(df_w/dS_w)_{S_{wf}}\approx2.747$. In this linear, uniform-cross-section "shoestring" system the whole width is flooded uniformly, so the areal sweep efficiency equals the fractional distance the front has advanced, $x_{D}=(df_w/dS_w)_{S_{wf}}\,Q_{iD}$:
$$Q_{iD}=\frac{25{,}000}{160{,}285}=0.1560\text{ PV}$$
$$x_D=2.747\times0.1560$$
$$\boxed{\text{Areal sweep}\approx 42.8\%\ (\text{pre-breakthrough; breakthrough occurs at }Q_{iD}=1/2.747=0.364\text{ PV})}$$
